# 1963 AHSME Problems/Problem 35

## Problem

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is $21$ and the perimeter is $48$. The shortest side is: $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$

## Solution

Let $b$ and $c$ be the other two sides of the triangle. The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.

By Heron's Formula, the area of the triangle is $\sqrt{24 \cdot 3(24-b)(b-3)}$. Plug in the answer choices for $b$ and write the prime factorization of the product to make sure it is a perfect square.

Testing $b = 8$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 16 \cdot 5} = \sqrt{2^7 \cdot 3^2 \cdot 5}$, so $8$ does not work. However, testing $b = 10$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}$, so $10$ works. The third side is $17$, and the sides satisfy the Triangle Inequality, so the answer is $\boxed{\textbf{(B)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 